Solving the Projectile and Ramp Problem

[physics_noob2]

Homework Statement

A projectile starts from rest and moves 4.7 m down a frictionless ramp inclined at 21 degrees with the horizontal. the acceleration due to gravity is 9.8 m/s^2

what what speed will it leave the ramp ?

what will the range of the projectile if the bottom of the ramp is 2.1 m above the ground

Homework Equations

V=Vi + at
delta X = Vi*t+1/2 a*t^2
Vf^2 - Vi^2 = 2*a*deltaX

The Attempt at a Solution

Vertical,
Viy=0
ay=-9.8
Vfy= - Vf sin 21
delta y= -4.7 sin 21

horizontal,
Vix= ( i thought it to be zero at first but it can't be zero because then the delta x comes out to be zero. So I am confused at this one )
ax=0
Vfx=Vf cos 21
delta x= 4.7 cos 21

so i used the third law and set it up as follows,

Vfy^2 = 2 (-9.8) (4.7 sin 21)

then the idea is to set the answer = Vf sin 21 and solve for Vf

it comes out to be 16.0329

but this is unfortunately not the right answer. can anyone help ?

 

[alphysicist]

Hi physics_noob2,

Homework Statement

A projectile starts from rest and moves 4.7 m down a frictionless ramp inclined at 21 degrees with the horizontal. the acceleration due to gravity is 9.8 m/s^2

what what speed will it leave the ramp ?

what will the range of the projectile if the bottom of the ramp is 2.1 m above the ground

Homework Equations

V=Vi + at
delta X = Vi*t+1/2 a*t^2
Vf^2 - Vi^2 = 2*a*deltaX

The Attempt at a Solution

Vertical,
Viy=0
ay=-9.8

I don't believe this is true; while it is on the ramp, there is another force besides gravity acting on the object.

More importantly, while it is on the ramp it is only moving in a single direction, so this part of the problem only has one-dimensional motion. What is the acceleration in the direction of motion? Once you have that, you can find the velocity at the end of the ramp.

Vfy= - Vf sin 21
delta y= -4.7 sin 21

horizontal,
Vix= ( i thought it to be zero at first but it can't be zero because then the delta x comes out to be zero. So I am confused at this one )
ax=0
Vfx=Vf cos 21
delta x= 4.7 cos 21

so i used the third law and set it up as follows,

Vfy^2 = 2 (-9.8) (4.7 sin 21)

then the idea is to set the answer = Vf sin 21 and solve for Vf

it comes out to be 16.0329

but this is unfortunately not the right answer. can anyone help ?